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If LPG cylinder contains mixture of buta...

If LPG cylinder contains mixture of butane and isobutane, then the amount of oxygen that would be required for combustion of 1 kg of it will be

A

`1*8 kg`

B

`2*7 kg`

C

`4*5 kg`

D

`3*58 kg`

Text Solution

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The correct Answer is:
To determine the amount of oxygen required for the combustion of 1 kg of a mixture of butane (C4H10) and isobutane, we will follow these steps: ### Step 1: Write the balanced chemical equation for the combustion of butane. The balanced equation for the combustion of butane is: \[ \text{C}_4\text{H}_{10} + \frac{13}{2} \text{O}_2 \rightarrow 4 \text{CO}_2 + 5 \text{H}_2\text{O} \] ### Step 2: Calculate the molar mass of butane (C4H10). The molar mass of butane can be calculated as follows: - Carbon (C): 12 g/mol × 4 = 48 g/mol - Hydrogen (H): 1 g/mol × 10 = 10 g/mol - Total molar mass of butane = 48 g/mol + 10 g/mol = 58 g/mol ### Step 3: Determine the amount of oxygen required for the combustion of butane. From the balanced equation, we see that 1 mole of butane requires \(\frac{13}{2}\) moles of oxygen for complete combustion. ### Step 4: Calculate the molar mass of oxygen (O2). The molar mass of oxygen (O2) is: - Oxygen (O): 16 g/mol × 2 = 32 g/mol ### Step 5: Calculate the mass of oxygen required for the combustion of 58 g of butane. Using the stoichiometry from the balanced equation: - 1 mole of butane (58 g) requires \(\frac{13}{2}\) moles of oxygen. - Mass of oxygen required = \(\frac{13}{2} \times 32\) g = 208 g ### Step 6: Calculate the amount of oxygen required for 1 kg of butane. Since we have calculated the oxygen required for 58 g of butane, we can find the amount required for 1 kg (1000 g): - Oxygen required for 1 kg of butane = \(\frac{208 \text{ g}}{58 \text{ g}} \times 1000 \text{ g}\) Calculating this gives: \[ \text{Oxygen required} = \frac{208}{58} \times 1000 \approx 3586.21 \text{ g} \] Converting grams to kilograms: \[ \text{Oxygen required} \approx 3.586 \text{ kg} \] ### Conclusion: The amount of oxygen required for the combustion of 1 kg of the mixture of butane and isobutane is approximately **3.58 kg**. ---

To determine the amount of oxygen required for the combustion of 1 kg of a mixture of butane (C4H10) and isobutane, we will follow these steps: ### Step 1: Write the balanced chemical equation for the combustion of butane. The balanced equation for the combustion of butane is: \[ \text{C}_4\text{H}_{10} + \frac{13}{2} \text{O}_2 \rightarrow 4 \text{CO}_2 + 5 \text{H}_2\text{O} \] ### Step 2: Calculate the molar mass of butane (C4H10). The molar mass of butane can be calculated as follows: ...
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    B
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  • What is the weight of oxygen that is required for the complete combustion of 2.8 kg of ethylene

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    B
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    C
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    D
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